Volume 10 (2009), Issue 1, Article 10, 5 pp.
A GENERAL INEQUALITY OF NGÔ-THANG-DAT-TUAN TYPE
TAMÁS F. MÓRI
D EPARTMENT OF P ROBABILITY T HEORY AND S TATISTICS
L ORÁND E ÖTVÖS U NIVERSITY
PÁZMÁNY P. S . 1/C, H-1117 B UDAPEST, H UNGARY
moritamas@ludens.elte.hu
Received 04 November, 2008; accepted 14 January, 2009
Communicated by S.S. Dragomir
A BSTRACT. In the present note a general integral inequality is proved in the direction that was
initiated by Q. A. Ngô et al [Note on an integral inequality, J. Inequal. Pure and Appl. Math.,
7(4) (2006), Art.120].
Key words and phrases: Integral inequality, Young inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In their paper [7] Ngô, Tang, Dat, and Tuan proved the following inequalities. If f is a
nonnegative, continuous function on [0, 1] satisfying
Z 1
Z 1
f (t) dt ≥
t dt, ∀x ∈ [0, 1],
x
x
then
Z
1
α+1
f (x)
0
Z
dx ≥
1
α
1
Z
α+1
x f (x) dx,
0
f (x)
0
Z
dx ≥
1
x f (x)α dx
0
for every positive number α.
This result has initiated a series of papers containing various extensions and generalizations
[1, 2, 3, 5, 6]. Among others, it turns out that the conditions above imply
Z 1
Z 1
α+β
f (x)
dx ≥
xα f (x)β dx
0
0
for every α > 0, β ≥ 1, which answered an open question of Ngô et al. in the positive [3].
The aim of this note is to formulate and prove a further generalization. It is presented in
Section 2. Section 3 contains corollaries, which are immediate extensions of a couple of known
results.
This research has been supported by the Hungarian National Foundation for Scientific Research, Grant No. K 67961.
298-08
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TAMÁS F. M ÓRI
2. M AIN R ESULT
Theorem 2.1. Let u, v : [0, +∞) → R be nonnegative, differentiable, increasing functions.
0
Suppose that u0 (t) is positive and increasing, and v u(t)u(t)
is increasing for t > 0. Let f and g
0 (t)
be nonnegative, integrable functions defined on the interval [a, b]. Suppose g is increasing, and
Z b
Z b
(2.1)
g(t) dt ≤
f (t) dt
x
x
holds for every x ∈ [a, b]. Then
Z b
Z b
Z b
(2.2)
u(g(t))v(g(t)) dt ≤
u(f (t))v(g(t)) dt ≤
u(f (t))v(f (t)) dt,
a
a
a
Z b
Z 1
(2.3)
u(g(t))v(f (t)) dt ≤
u(f (t))v(f (t)) dt,
a
0
provided the integrals are finite.
Remark 1.
(1) Here and throughout, by increasing we always mean nondecreasing.
(2) Note that continuity of f or g is not required.
(3) Unfortunately, the other inequality
Z 1
Z 1
(2.4)
u(g(t))v(g(t)) dt ≤
u(g(t))v(f (t)) dt,
0
0
which seems to be missing from (2.3), is not necessarily valid. Set [a, b] = [0, 1],
u(t) = tβ , v(t) = tα , with α > 0, β > 1. Let g(t) = t, and f (t) = 1, if 1/2 ≤ t ≤ 1,
and zero otherwise. Then all the conditions of Theorem 2.1 are satisfied, and
Z b
Z 1
1
,
u(g(t))v(g(t)) dt =
tα+β dt =
α+β+1
a
0
Z b
Z 1
1
1
β
u(g(t))v(f (t)) dt =
t dt =
1 − β+1 .
β+1
2
a
1/2
It is easy to see that (2.4) does not hold if α <
β+1
.
2β+1 −1
Although f is discontinuous in this counterexample, it is not continuity that can help, for f
can be approximated in L1 with continuous (piecewise linear) functions.
For the proof we shall need the following lemmas of independent interest.
Lemma 2.2. Let f and g be nonnegative integrable functions on [a, b] that satisfy (2.1). Let
h : [a, b] → R be nonnegative and increasing. Then
Z b
Z b
(2.5)
h(t)g(t) dt ≤
h(t)f (t) dt.
a
a
Proof. We can suppose that u is right continuous, because it can only have countably many discontinuities, soRreplacing u(t) with u(t+) in these points does not affect the integrals. Clearly,
h(t) = h(a) + (a,t] dh(s), hence
Z t+
Z b
Z b
h(t)g(t) dt =
h(a) +
dh(s) g(t) dt
a
a
a+
Z
= h(a)
b
Z bZ
a
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 10, 5 pp.
b
I(s ≤ t)g(t) dh(s) dt,
g(t) dt +
a
a
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where I(·) stands for the characteristic function of the set in brackets. By Fubini’s theorem we
can interchange the order of the integration, obtaining
Z b
Z b
Z bZ b
h(t)g(t) dt = h(a)
g(t) dt +
I(s ≤ t)g(t) dt dh(s)
a
a
a
a
Z b
Z b Z b
= h(a)
g(t) dt +
g(s) ds dh(s).
a
a
t
Remembering condition (2.1), we can write
Z b
Z b
Z b Z b
h(t)g(t) dt ≤ h(a)
f (t) dt +
f (s) ds dh(s)
a
a
Z
=
a
t
b
h(t)f (t) dt,
a
as required.
Lemma 2.3. Let f and g be as in Theorem
2.1, and let v : [0, +∞) → R be a nonnegative
Rx
increasing function. Define V (x) = 0 v(t) dt, x ≥ 0. Then
Z b
Z b
(2.6)
V (g(t)) dt ≤
V (f (t)) dt.
a
a
Equivalently, we can say that inequality (2.6) is valid for all increasing convex functions
V : [0, +∞) → R.
Proof. We can suppose that the right-hand side is finite, for the integrand on
R xthe left-hand side
is bounded. Let V ∗ denote the Legendre transform of V , that is, V ∗ (x) = 0 v −1 (t) dt, where
v −1 (t) = inf{s : v(s) ≥ t} is the (right continuous) generalized inverse of v. Then by the
Young inequality [4] we have that xy ≤ V (x) + V ∗ (y) holds for every x, y ≥ 0, with equality
if and only if v(x−) ≤ y ≤ v(x+). Hence, by substituting x = f (t) and y = v(g(t)) we obtain
(2.7)
f (t)v(g(t)) ≤ V (f (t)) + V ∗ (v(g(t))) = V (f (t)) + g(t)v(g(t)) − V (g(t)).
By integrating this we get that
Z b
Z b
Z b
Z b
(2.8)
f (t)v(g(t)) dt ≤
V (f (t)) dt +
g(t)v(g(t)) dt −
V (g(t)) dt.
a
a
a
With h(t) = v(g(t)) Lemma 2.2 yields
Z b
Z
(2.9)
g(t)v(g(t)) dt ≤
a
a
1
f (t)v(g(t)) dt.
0
Combining (2.8) with (2.9) we arrive at (2.6).
Proof of Theorem 2.1. First we prove for the case where u(t) = t. Then t v 0 (t) has to be increasing.
The first inequality of (2.2) has already been proved in (2.9). On the other hand, from the
Young inequality, similarly to (2.7) we can derive that
f (t)v(g(t)) ≤ V (f (t)) + V ∗ (v(g(t)))
= V ∗ (v(g(t))) + f (t)v(f (t)) − V ∗ (v(f (t))).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 10, 5 pp.
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4
TAMÁS F. M ÓRI
Therefore,
Z b
Z b
Z b
Z b
∗
(2.10)
f (t)v(g(t)) dt ≤
V (v(g(t))) dt +
f (t)v(f (t)) dt −
V ∗ (v(f (t))) dt.
a
a
a
Here
Z
∗
V (v(x)) = xv(x) − V (x) =
x
a
0
(tv(t)) − v(t) dt =
Z
0
x
tv 0 (t) dt,
0
thus Lemma 2.3 can be applied with V ∗ (v(x)) in place of V (x).
Z b
Z b
∗
(2.11)
V (v(g(t))) dt ≤
V ∗ (v(f (t))) dt.
a
a
Now we can complete the proof of the second inequality of (2.2) by plugging (2.11) back into
(2.10).
Next, since [f (t) − g(t)][v(f (t)) − v(g(t))] ≥ 0, we obtain that
Z 1
Z 1
Z 1
Z 1
f (t)v(f (t)) dt −
g(t)v(f (t) dt ≥
f (t)v(g(t)) dt −
g(t)v(g(t) ≥ 0,
0
0
0
0
by (2.2). This proves (2.3).
For the general case, we first apply Lemma 2.3 on the interval [x, b], with u(t) in place of
V (t). We can see that u(f (t)) and u(g(t)) satisfy condition (2.1). Now, u is invertable. Let
w(t) = v(u−1 (t)), then
v 0 (u−1 (t))
,
w0 (t) = 0 −1
u (u (t))
hence, by the conditions of Theorem 2.1, tw0 (t) is increasing. The proof can be completed by
applying the particular case just proved to the functions u(f (t)) and u(g(t)), with w in place of
v.
3. C OROLLARIES , PARTICULAR C ASES
In this section we specialize Theorem 2.1 to obtain some well known results that were mentioned in the Introduction. First, let u(x) = xβ and v(x) = xα with α > 0 and β ≥ 1. They
clearly satisfy the conditions of Theorem 2.1.
Corollary 3.1. Let f and g be nonnegative, integrable functions defined on the interval [a, b].
Suppose g is increasing, and
Z b
Z b
(3.1)
g(t) dt ≤
f (t) dt
x
x
holds for every x ∈ [a, b]. Then, for arbitrary α > 0 and β ≥ 1 we have
Z b
Z b
Z b
α+β
α
β
(3.2)
g(t)
dt ≤
g(t) f (t) dt ≤
f (t)α+β dt,
a
a
a
Z b
Z b
(3.3)
f (t)α g(t)β dt ≤
f (t)α+β dt.
a
a
Next, change α, β, f (t), and g(t) in Corollary 3.1 to α/β, 1, f (t)β and g(t)β , respectively.
Corollary 3.2. Let α and β be arbitrary positive numbers. Let f and g satisfy the conditions of
Corollary 3.1, but, instead of (3.1) suppose that
Z b
Z b
β
(3.4)
g(t) dt ≤
f (t)β dt
x
x
holds for every x ∈ [a, b]. Then inequalities (3.2) and (3.3) remain valid.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 10, 5 pp.
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In particular, for the case of [a, b] = [0, 1], g(t) = t Corollary 3.1 yields Theorem 2.3 of [3],
and Corollary 3.2 implies Theorem 2.1 of [5]. If, in addition, we set β = 1, Corollary 3.1 gives
Theorems 3.2 and 3.3 of [7].
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J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 10, 5 pp.
http://jipam.vu.edu.au/